1. ## Partial Fractions

I'm having trouble setting up the partial fractions for this integral

$\int{\frac{x^2-3x+7}{(x^2-4x+6)^2}}$

I think this is the case of a repeating, irreducible quadratic factor. So, would the decomposition be this?

$\frac{A}{(x^2-4x+6)} + \frac{B}{(x^2-4x+6)^2}$

2. Originally Posted by centenial
I'm having trouble setting up the partial fractions for this integral

$\int{\frac{x^2-3x+7}{(x^2-4x+6)^2}}$

I think this is the case of a repeating, irreducible quadratic factor. So, would the decomposition be this?

$\frac{A}{(x^2-4x+6)} + \frac{B}{(x^2-4x+6)^2}$
Note quite remember when you have a quadratic in the denominator the numerator have the form

$Ax+B$ so it should look like this

$\displaystyle \frac{Ax+B}{(x^2-4x+6)} + \frac{Cx+D}{(x^2-4x+6)^2}$

3. Thanks so much!

4. Just in case you also want to do it by 'the method of avoiding partial fractions':

Observe that $(x^2-3x+7)-(x^2-4x+6)-(x+1) = 0.$ Then:

Spoiler:
$\displaystyle \int{\frac{x^2-3x+7}{(x^2-4x+6)^2}}\;{dx} = \int{\frac{(x^2-3x+7)-(x^2-4x+6)-(x+1)+(x^2-3x+7)}{(x^2-4x+6)^2}}\;{dx}$
Spoiler:

$\displaystyle = \int{\frac{(x^2-3x+7)-(x^2-4x+6)}{(x^2-4x+6)^2}}\;{dx}-\int{\frac{(x+1)-(x^2-3x+7)}{(x^2-4x+6)^2}}\;{dx}$

Spoiler:
$\displaystyle = \int{\frac{x^2-3x+7-x^2+4x-6}{(x^2-4x+6)^2}}\;{dx}-\int{\frac{x+1-x^2+3x-7}{(x^2-4x+6)^2}}\;{dx}$

Spoiler:
$\displaystyle = \int{\frac{x+1}{(x^2-4x+6)^2}}\;{dx}-\int{\frac{-(x^2-4x+6)}{(x^2-4x+6)^2}}\;{dx}$

Spoiler:
$\displaystyle = \int{\frac{x+1}{(x^2-4x+6)^2}}\;{dx}+\int{\frac{1}{(x^2-4x+6)}}\;{dx}$
Spoiler:
You can do your usual integration (completing the square etc) now.

5. Originally Posted by centenial
I'm having trouble setting up the partial fractions for this integral

$\int{\frac{x^2-3x+7}{(x^2-4x+6)^2}}$

I think this is the case of a repeating, irreducible quadratic factor. So, would the decomposition be this?

$\frac{A}{(x^2-4x+6)} + \frac{B}{(x^2-4x+6)^2}$
Or....denominator is $x^2-4x+6$ squared.

If we add an x and a 1, we will have the numerator...

$\displaystyle\frac{\left(x^2-4x+6\right)+(x+1)}{\left(x^2-4x+6\right)^2}=\frac{1}{x^2-4x+6}+\frac{x+1}{\left(x^2-4x+6\right)^2}$

6. Originally Posted by Archie Meade
Or....denominator is $x^2-4x+6$ squared.

If we add an x and a 1, we will have the numerator...

$\displaystyle\frac{\left(x^2-4x+6\right)+(x+1)}{\left(x^2-4x+6\right)^2}=\frac{1}{x^2-4x+6}+\frac{x+1}{\left(x^2-4x+6\right)^2}$
Looks like a zero term made me do works!